A problem by GH Hardy

The pattern is that the denominator increases from 1, and the numerator decreases from some integer, $n.$ Then the process repeats with $n+1.$

${\color{#D61F06}\frac 1{1}},\ {\color{#3D99F6}\frac {2}{1},\ \frac {1}{2}},\ {\color{#20A900}\frac {3}{1},\ \frac {2}{2},\ \frac {1}{3}},\ {\color{#69047E}\frac {4}{1},\ \frac {3}{2},\ \frac {2}{3},\ \frac {1}{4}},\ \ldots$

Since the denominator increases by 1 while the numerator decreases by 1, the same-colored fractions always have the same sum of numerator and denominator.

Incidentally, the fractions we are given have the same sum of numerator and denominator:

$\begin{aligned} \frac{1887}{1920} &: 1887+1920=3807 \\ \\ \frac{1789}{2018} &: 1789+2018=3807 \end{aligned}$

Since these sums are the same, they are in the same sub-sequence (as if they were the same color). Then the difference between their indices is $2018-1920=98.$ And so $\mid m-n \mid = \boxed{98}.$

Incidentally, 1887 and 1920 are the birth-year and death-year of mathematician Srinivasa Ramanujan.

The sequence consists of infinitely many subsequences in order:

• 1 fraction where the sum of the numbers equals 2,

• 2 fractions where the sum of the numbers equals 3,

• 3 fractions where the sum of the numbers equals 4,

• ...

• $s - 1$ fractions where the sum of the numbers equals $s$ ,

• ...

Since $1887 + 1920 = 1789 + 2018$ , the two given fractions are part of the same subsequence.

Now note that within each subsequence with sum $s$ , the denominators increase $1, 2, \dots, s-1$ while the numerators decrease $s-1, s-2, \dots, 1$ . Since each denominator is one more than the previous one, the distance between two fractions in a subsequence is equal to the difference of their denominators.

In this case, $\text{distance} = |m-n| = |2018 - 1920| = \boxed{98}.$

The list of numerators and denominators can both be shown in their respective triangular list, like this: We want to figure out their position in the sequence: (Note that the triangle numbers are on the right-end of each row.)

Let $N$ be the numerator, $D$ be the denominator, and $R$ be the row of the $n^{th}$ term in the sequence.

From the triangle of denominators, we can see that each term is $D$ places after the end of the previous row $(R-1)$ . Using the triangle number formula $\frac{n^2+n}{2}$ , we can find the position from $\frac{(R-1)^2+(R-1)}{2}+D$ .

Now we need to find the row of the $n^{th}$ term. To do this, consider the sums of each numerator-denominator pair: they all result in the same distinct numbers for each row. With this, we can find the row for each term using $R=N+D-1$ . Substituting this into the formula above gives us:

$f(n)=\frac{(N+D-2)^2+(N+D-2)}{2}+D$

Plugging $\frac{1887}{1920}$ and $\frac{1789}{2018}$ into this formula, we get $7242835$ and $7242933$ respectively. With these, we can get our final answer: $|7242835-7242933|=\boxed{98}$

분모 + 분자 = 3807 나오는데, 시그마 n 공식인 n(n+1)/2에서 m번째와 n번째를 찾으면, {3806(3807)/2 + 1887} - {3806(3807)/2 + 1789} = 1887-1789 = 98

Premise : I'm not English so I'm sorry for any mistake in the text.

First Solution :

1887+1920=1789+2018 => solution: |2018-1920|=98

Second solution :

$|(\sum_{x=a+b}^{c+d-1} x)-((c+d)-(a+b))-(b-1)+(d-1)| = |(\sum_{x=a+b}^{c+d-1} x)+a-c|$

As $\frac{a}{b} = \frac{1887}{1920}$ ; $\frac{c}{d} = \frac{1789}{2018}$

Number of jumps performed from the fraction in the serie with denominator equivalent to 1 and sum of numerator and denominator equivalent to (a+b) to the other with same denominator but sum equivalent to (c+d):

$(\sum_{x=a+b}^{c+d-1} x)$

Number of unnecessary jumps performed in the precedent step:

$((c+d)-(a+b))$

Offset from the first equation with denominator 1 to the actual fraction $\frac{a}{b}$

$(b-1)$

Offset from the second equation with denominator 1 to the actual fraction $\frac{c}{d}$

$(d-1)$

$|(\sum_{x=a+b}^{c+d-1} x)+a-c| = 98$

Since denominator+numerator is the same in both terms, the solution simply is 1887-1789=||98||.

The sequence of the numerator can be modeled by $a(x)=\frac{x(x+1)}{2} -x+1$ , where $x$ is the value of the numerator and $a(x)$ is equal to the first term at which that value appears in the sequence (where $x$ and $a(x)$ are positive integers).

The sequence of the denominator can be modeled by $b(x)=\frac{x(x+1)}{2}$ , where $x$ is the value of the denominator and $b(x)$ is equal to the first term at which that value appears in the sequence (where $x$ and $b(x)$ are positive integers).

Now we must account for the fact every positive integer appears multiple times in the sequence. We must write functions to describe the second, third, etc. term that a given value appears in the sequence.

We can account for this in $a(x)$ with the function $a_{i}(x)=a(x+i)-x$ , where $i$ is a positive integer that describes if this is the first ( $i=1$ ), second ( $i=2$ ), third ( $i=3$ ), etc. term at which the value $x$ appears in the sequence.

We account for this in $b(x)$ with the function $b_{j}(x)=b(x+j-2)+x$ , where $j$ is a positive integer that describes if this is the first ( $j=1$ ), second ( $j=2$ ), third ( $j=3$ ), etc. term at which the value $x$ appears in the sequence.

To first find $m$ , we plug in our values $1887$ and $1920$ into $a_{i}(x)$ and $b_{j}(x)$ respectively. The term at which $a_{i}(1887)$ = $b_{j}(1920)$ is equal to $m$ .

We find that $a_{i}(1887)=b_{j}(1920)$ at $i=1920$ , $j=1887$ .

$a_{1920}(1887)$ and $b_{1887}(1920)$ are both equal to $7242835$ , which is the value of $m$ .

Following the same set of steps for $\frac{1789}{2018}$ to find $n$ , we find that $a_{i}(1789)=b_{j}(2018)$ at $i=2018$ , $j=1789$ .

$a_{2018}(1789)$ and $b_{1789}(2018)$ are both equal to $7242933$ , which is the value of $n$ .

Plugging our values for $m$ and $n$ into $|m-n|$ , we find that $|7242835-7242933|=98$ .

(m)th no is 1887/1920; (n)th is 1789/2018 here( 2018-1920)=(1887-1789)=98 so started from same number. 1887+1919=3806; If not started from same number

Let as represent them as {A/B} and {C/D},

X)find position of (A/1) as[{(A-1)÷2}×A]+1 【it is the sum of all numbers till (A-1)+1】

●For1887/1920,position of 1887/1 is (1886÷2)×1887+(1)=1779442.from the start ●For 1789/2018,position of 1789/1 is (1788÷2)×1789+(1)=1599367 from the start

Y)find position of A/B from A/1 {A×(B-1)}+[{(B-1)÷2}×B]

●For 1887/1920 it is (1887×1919)+(1919÷2)×1920= 3621153+1842240=5463393 ●For 1789/2018 it is (1789×2017)+(2017÷2)×2018= 3608413+2035153=5643566

So the position of A/B is X+Y from the start,

●For 1887/1920 it is 1779442+5463393=7242835 from start ●For 1789/2018 it is 1599367+5643566=7242933 from start

So the difference is 7242933-7242835=98

Visualising Pascal's triangle makes this much easier. First, calculating the m^th term, 1887/1920. As we know 1/1920 is on the 1920th row, and to get to the row that 1887/1920, you have to add 1887-2 to the denominator, because you dont count the row 1/1920 is on and the row it is actually on. So 1885+1920=3805. Do the same thing for the other number and you also get 3805. Then you the calculate the difference between the two numerators which is -98(m-n), sub into 1/2|(m-n)|, which results in 49

The denominator reset to 1 in every power of 2 going down from the value of that power, so a given number is in a subsequence if their integer value of its log2 is the same. log2(1920) = 10.9068 -> 10 log2(2018) = 10.9787 -> 10 M is bigger than N because is closer to 1

Then their difference would be the same in whichever subsequence they are. In this case, the difference is -98, but being the "lowest" number in a greater position, this number should be taken as positive. Divided by 2 gives us 49 as result.